Respuesta :
Answer:
The object hits the ground in 5 seconds.
Step-by-step explanation:
Given : An object is launched at 9.8 meters per second from 73.5 meter tall platform. The objects height s (in meters) after t seconds I’d given by the equation [tex]s(t)= -4.9t^2-9.8t+73.5[/tex].
To find : Use the factoring to determine when the object hits the ground ?
Solution :
When the object hits the ground the distance is zero.
i.e. s(t)=0
So, [tex]-4.9t^2-9.8t+73.5=0[/tex]
Applying quadratic formula, [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Here, a=-4.9, b=-9.8 and c=73.5
[tex]t=\frac{-(-9.8)\pm\sqrt{(-9.8)^2-4(-4.9)(73.5)}}{2(-4.9)}[/tex]
[tex]t=\frac{9.8\pm\sqrt{1536.64}}{-9.8}[/tex]
[tex]t=\frac{9.8\pm 39.2}{-9.8}[/tex]
[tex]t=1\pm 4[/tex]
[tex]t=1+4,1-4[/tex]
[tex]t=5,-3[/tex]
Reject t=-3
Therefore, the object hits the ground in 5 seconds.
